Solving Equations with 2 Variables
Equations with two variables have two unknowns in one equation, like x + 2y = 20. You solve it by substituting a value for one variable to find the other, so x = 10 gives y = 5. They have many solution pairs, not just one.
Video Lesson
Watch and learn the basics
🎬 Did this video explain it clearly?
Flashcards
Review key concepts visually
%20Equations%20with%202%20Variables.webp)
%20Solving%20Equation%20with%202%20variables.webp)
%20Solving%20Equations%20with%202%20Variables%20Graphically.webp)
Equations with Two Variables
- An equation like has two unknowns.
- There are many solutions because lots of number pairs can work.
Solving by Substitution
- Substitute the value you know into the equation to solve for the other variable.
- For example, if , substitute it in to get . Solving for y, we get .
Solving Graphically
- Rewrite the equation as to draw the line.
- This means every point on the line is a solution to the equation.
- The point (8, 6) is a solution because it lies on the line.
Practice Questions
Test your understanding
You have £20. Each snack costs £2, and each water costs £1. Use x for the number of snacks and y for the number of waters. How would you form the equation?
Correct! 🎉 +10 pointsNot quite right
The cost of x snacks is , and the cost of y waters is y. The total amount is 20, so the equation is .
You have £30. Each apple costs £3, and each orange costs £2. Use x for the number of apples and y for the number of oranges. How would you form the equation?
Correct! 🎉 +10 pointsNot quite right
The cost of x apples is , and the cost of y oranges is . The total amount is 30, so the equation is .
You have £40. Each pencil costs £2, and each eraser costs £3. If you buy 5 pencils, how many erasers can you still afford?
Correct! 🎉 +20 pointsNot quite right
The equation formed here is , where x is the number of pencils and y is the number of erasers. Buying 5 pencils gives us . Solving this, we find . So, you can still afford 10 erasers.
You have £40. Each pencil costs £2, and each eraser costs £3. If you buy 8 erasers, how many pencils can you still afford?
Correct! 🎉 +20 pointsNot quite right
The equation formed here is , where x is the number of pencils and y is the number of erasers. Buying 8 erasers gives us , which simplifies to . Solving this, we find . So, you can still afford 8 pencils.
You have £40. Each pencil costs £2, and each eraser costs £3. If you buy 6 pencils and 3 erasers, how much money will you have left?
Correct! 🎉 +20 pointsNot quite right
The equation formed here is , where x is the number of pencils and y is the number of erasers. Buying 6 pencils and 3 erasers gives us . So, you've spent 21. Subtracting this from your total, , meaning you have 19 left.
You have £40. Each pencil costs £2, and each eraser costs £3. In the graph below, what does point P mean?
Correct! 🎉 +30 pointsNot quite right
The equation represented is , where x is the number of pencils and y is the number of erasers. Point P shows and , meaning 5 pencils and 10 erasers can be purchased.
Want to see the full working?
Interactive Activity
Solving equations with 2 variables
Loading interactive widget...
Students Also Ask
The questions students bump into most on this topic
Substitute a value for one variable, then solve for the other. For example, in x + 2y = 20, putting x = 10 gives 10 + 2y = 20, so 2y = 10 and y = 5. You can also put in a value for y to find x.
A linear equation in two variables has many solutions, not just one. Lots of different x and y pairs can make it true. For x + 2y = 20, both (10, 5) and (4, 8) work, so the equation is satisfied by many value pairs.
A simple example is x + 2y = 20, which models spending £20 on cookies at £1 each (x) and sandwiches at £2 each (y). It has two variables, x and y, and many solution pairs, such as x = 10 with y = 5.
First rearrange the equation to make y the subject. For x + 2y = 20, this gives y = 10 - 0.5x, with a y-intercept of 10 and a gradient of -0.5. Plot that straight line, and every point on it is a solution pair.
Because two unknowns can change together. When you choose a value for one variable, you can always work out a matching value for the other. Each valid choice gives a new pair, so the equation has many solutions, shown as all the points along its line.